Question

The integral ${\int }_0^{1/2}\frac{\ln (1+2x)}{1+4x^2}dx$ equals :

• A. $\frac{\pi }{4}ln2$
• B. $\frac{\pi }{8}ln2$
• C. $\frac{\pi }{12}ln2$
• D. $\frac{\pi }{32}ln2$

Answer 1

The correct option is C $\frac{\pi }{12}ln2$

$I$=${\int }_0^{1/2}\frac{\log (1+2x)}{1+4x^2}dx$
Let $2x=\tan \theta$

$2dx$$={\sec }^2\theta d\theta$
$I={\int }_0^{\pi /4}\frac{1}{2}\log (1+\tan \theta )d\theta$
$I=\frac{1}{2}{\int }_0^{\pi /4}\log (1+\tan (\frac{\pi }{4}-\theta \left)\right)d\theta$
$=\frac{1}{2}{\int }_0^{\pi /4}\log \left(\frac{2}{1+\tan \theta }\right)d\theta$
$=\frac{1}{2}{\int }_0^{\pi /4}(\log 2-\log (1+\tan \theta \left)\right)d\theta$
$\frac{3}{2}I=\frac{1}{2}{\int }_0^{\pi /4}\log 2d\theta$
$I=\frac{\pi }{12}log2$